In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid into enough distilled water to make 500.0 mL of solution. 30.0 mL of the base was measured, and in the titration, 16.74 mL of the acid titrant was needed to neutralize. Determine the concentration of the Sodium hydroxide.

1.395M NaOH

Explanation:

Sodium hydroxide, NaOH, reacts with nitric acid, HNO3, as follows:

NaOH + HNO3 → NaNO3 + H2O

Where 1mol of NaOH reacts with 1mol of HNO3

To solve this question we must find the concentration of the titrant. With the concentration and the needed acid we can find the moles of HNO3 added = moles NaOH in the solution. With the moles of NaOH and its volume we can find its concentration as follows:

HNO3 concentration:

10.00mol/L HNO3 * (125.0mL/500.0mL) = 2.500M HNO3

Moles HNO3 = Moles NaOH:

16.74mL = 0.01674L*(2.500mol/L) = 0.04185 moles HNO3 = Moles NaOH

Concentration NaOH:

0.04185 moles / 0.0300L =

1.395M NaOH

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell